Probability of the Union of Non- Mutually Exclusive Events

The probability of the union of two events A and B written as P(AUB) or P(A or B) is equal to sum of the probability of event A, P(A) and the probability of event B, P(B) minus the probability of events A and B occurring together (P(A∩B). In symbols,

P(AUB) = P(A) + P(B) - P(A∩B)

Example 1: If a card is drawn from an ordinary deck of 52 cards,

find the probability of getting a heart or a face card?

P(AUB) = P(A) + P(B) - P(A∩B)

P(heart or face card) = P(heart) + P(face card) – P(heart n face card)

P(heart)= 13/52

Pface card) = 12/52

P(heart n face card) = 3/52

P(heart or face card) = 13/52 + 12/52 - 3/52 = 22/52 =11/26

Example 2: A card is drawn at

random from a 52 - deck of

card. Find the Probability of

getting:

a. a number3 or a heart?

b. a spade or a jack!

c. a black card or 10 of spades?

d. a face card or a diamond?

e. a queen or a spade?

Solution:

a. a number 3 or a heart?

4/52 + 13/52 - 1/52 = 16/52 = 4/13

b. a spade or a jack!

4/52 + 13/52 - 1/52 = 16/52 = 4/13

c. a black card or 10 of spades?

26/52 + 1/52 - 1/52 = 26/52 = 1/2

d. a face card or a diamond?

12/52 + 13/52 - 3/52 = 22/52 = 11/26

e. a queen or a spade?

4/52 + 13/52 - 1/52 = 16/52 = 4/13

Example 3: What is the probability that a number selected at random from the first 50 positive integers is exactly divisible by 3 or 4?

divisible by 3 = {3, 6, 9, 12 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45,48}

divisible by 4 = {4, 8, 12, 16, 20, 24, 28, 32,36, 40, 44, 48}

P(divisible by 3) = 16/50

P(divisible by 4) = 12/50

P(divisible by 3 ∩ divisible by 4) = 4/50

P(divisible by 3 U divisible by 4) = 16/50 + 12/50 - 4/50 = 4/24

 

Example 7: In a certain city, 60% of the residents watching either ABS-CBNbor GMA on the television. Thirty percent of them are watching ABS-CBN. Forty percent of them are watching GMA. What percentage of the residents are watching both channels?

Solution: 

Let A watching ABS-CBN be event A and watching GMA be event B?

P(AUB) = 0.6 P(A) = 0.3 P(B) = 0.4 P(A∩B)=?

P(AUB) = P(A) + P(B) - P(A∩B)

0.6 = 0.3 + 0.4- P(A∩B)

P(A∩B) = 0.3 + 0.4-0.6

P(A∩B) = 0.7-0.6

P(A∩B) = 0.1

Therefore, 10% of the population of a certain city arewatching both channels.